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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Cho a+b+c=0 và `a^2+b^2+c^2=2021`. Tính `a^4+b^4+c^4`

Toán Lớp 8: Cho a+b+c=0 và a^2+b^2+c^2=2021. Tính a^4+b^4+c^4

Comments ( 2 )

  1. $\\$
    a+b+c=0
    =>a^2+b^2+c^2+2(ab+bc+ca)=0
    => 2021+2(ab+bc +ca)=0
    => ab + bc +ca=(-2021)/2
    => (ab+bc+ca)^2 = 4084441/4
    => a^2b^2 + b^2c^2+c^2a^2 = 4084441/4
    a^2+b^2+c^2=2021
    =>(a^2+b^2+c^2)^2=4084441
    => a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4084441
    => a^4+b^4+c^4 + 2 . 4084441/4 = 4084441
    => a^4+b^4+c^4 + 4084441/2 = 4084441
    => a^4+b^4+c^4 =8168882/2 – 4084441/2
    =>a^4+b^4+c^4 =4084441/2=2042220,5
    Vậy a^4+b^4+c^4 =2042220,5
     

  2. a+ b+c=0
    <=> (a+b+c)^2 =0
    <=> a^2 + b^2 +c^2 + 2(ab +ac+ bc) =0
    <=> 2021 + 2(ab+ac+bc) =0
    <=> ab +ac +bc = -2021/2     
    <=> (ab+ac+bc)^2 = 4084441/4
    <=> (ab)^2 +(ac)^2 + (bc)^2 + 2abc(a+b+c) = 4084441/4
    <=> (ab)^2 +(ac)^2 + (bc)^2 + 2abc . 0 = 4084441/4
    <=> (ab)^2 +(ac)^2 + (bc)^2 =4084441/4
    Lại có :
    a^4 +b^4 +c^4 = (a^2 +b^2 +c^2)^2 -2[ (ab)^2 +(ac)^2 + (bc)^2]
    <=> a^4 +b^4 +c^4 = 2021^2 – 2 . 4084441/4 = 4084441/2  

    Vậy a^4 +b^4 +c^4 = 4084441/2 

     

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222-9+11+12:2*14+14 = ? ( )